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I asked this question on another forum but no answers so I'm copy/pasting it here in hopes that someone can help out

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  • $\begingroup$ Depending on $\text{sign}(x)$ the contour of integration is the upper/lower half plane. In the second case we have a pole and therefore a finite result, whereas in the first case the integrand is holomorpic inside the contour and therefore the ntegral is zero $\endgroup$ – tired Jun 26 '15 at 19:50
  • $\begingroup$ $x\mapsto e^{-ax}$ is for $a\in \mathbb{R}\setminus \{0\}$ not a tempered distribution, so it's unclear how you would take the Fourier transform of that. On the other hand, for $a > 0$, the function $x\mapsto e^{-ax}\cdot H(x)$ is a nice $L^1$-function whose Fourier transform is $\omega\mapsto \frac{1}{a+i\omega}$ (for the appropriate convention of the Fourier transform, slight modifications for the other conventions). $\endgroup$ – Daniel Fischer Jun 26 '15 at 19:53
  • $\begingroup$ Thanks for the answers, but they are just going over my head =/ $\endgroup$ – Maharero Jun 26 '15 at 20:10
  • $\begingroup$ The heavy use comes in whenever there are sign conventions that must preserved. $\endgroup$ – Zach466920 Jun 26 '15 at 20:34
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Try in the opposite direction: \begin{align} \mathcal{F}(e^{-ax}H(x)) &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-ax}H(x)e^{-isx}dx \\ & = \frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}e^{-ax}e^{-isx}dx \\ & = \frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}e^{-x(a+is)}dx \\ & = \frac{1}{\sqrt{2\pi}}\left.\frac{e^{-x(a+is)}}{a+is}\right|_{x=0}^{\infty} \\ & = \frac{1}{\sqrt{2\pi}}\frac{1}{a+is} \end{align} Because $\mathcal{F}^{-1}(\mathcal{F}(e^{-ax}H(x))=e^{-ax}H(x)$ (except at the discontinuity): $$ \mathcal{F}^{-1}\left(\frac{1}{\sqrt{2\pi}}\frac{1}{a+is}\right)=e^{-ax}H(x). $$ To see where $H(x)$ comes from using more general principles requires some background in Complex Analysis. If you like Fourier and Laplace transform analysis, you'd enjoy such a class. Complex Analysis is one of the most elegant subjects in Mathematics.

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